\newproblem{lay:5_4_18}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.4.18}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Define $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$ by $T(\mathbf{x})=A\mathbf{x}$, where $A$ is a $3\times 3$ matrix with eigenvalues 5, 5 and -2.
	Does there exist a basis $\mathcal{B}$ for $\mathbb{R}^3$ such that the $\mathcal{B}$-matrix of $T$ is a diagonal matrix? Discuss.
}{
  % Solution
	It depends on whether $A$ is diagonalizable or not. Since $A$ does not have all its eigen values distinct, the condition is (see Theorem 5.3.7) that the dimension
	of the eigenspace associated to eigenvalue 5 is 2, and that the dimension of the eigenspace associated to eigenvalue -2 is 1.
}
\useproblem{lay:5_4_18}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
